If $\pi: Z\rightarrow \mathbb{P}^n$ is a closed embedding where $Z$ is the zero set of some degree homogeneous polynomial, we have:
$$0\rightarrow \mathcal{O}_{\mathbb{P}^n}(-d)\rightarrow \mathcal{O}_{\mathbb{P}^n}\rightarrow \mathcal{O}_Z\rightarrow 0$$
where we think $\mathcal{O}_Z$ is really $\pi_*\mathcal{O}_Z$.
Because of this pushforward, I am confused how this short exact sequence interacts with tensoring with $\mathcal{O}_{\mathbb{P}^n}(1)$.
Isn't $\mathcal{O}_Z(1)$ defined to be $\pi^*\mathcal{O}_{\mathbb{P}^n}(1)$, does that make $(\pi_*\mathcal{O}_Z )\otimes \mathcal{O}_{\mathbb{P}^n}(m)$ equal to $\pi_*(\pi^*\mathcal{O}_{\mathbb{P}^n}(1))^{\otimes m}$, so we can say that
$$0\rightarrow \mathcal{O}_{\mathbb{P}^n}(-d+m)\rightarrow \mathcal{O}_{\mathbb{P}^n}(m)\rightarrow \mathcal{O}_Z(m)\rightarrow 0.$$
I am pretty confused about all this, but hopefully this makes enough sense that I can get some feedback.
Thank you for any help or advice
To get the second exact sequence from the 1st, just tensor up with $\mathcal{O}(m)$, we get $$0 \rightarrow \mathcal{O}(-d) \otimes \mathcal{O}(m) \rightarrow \mathcal{O} \otimes \mathcal{O}(m) \rightarrow (\pi_* \mathcal{O}_Z) \otimes \mathcal{O}(m) \rightarrow 0.$$Now, if I understand your question, you want to define $\mathcal{O}_Z(m)$ to be $\pi^* \mathcal{O}(m)$, so you want to know if $(\pi_* \mathcal{O}_Z) \otimes \mathcal{O}(m) \simeq \pi_* \pi^* \mathcal{O}(m).$ Here's a down to earth way to see this: if $Z$ was cut out by $f$, then on each standard affine chart $U_i$, both sides are just $k[U_i]/f(x_0, \ldots, 1, \ldots, x_n)$, where 1 is in the $i^{th}$ position, and both sides have the same transition functions from $U_i \rightarrow U_j$ (I'm using that you can glue coherent sheaves via identifications satisfying the cocycle condition, and that morphisms defined locally which are compatible on overlaps glue to a global morphism).